题解 | #统计每个用户的平均刷题数#

SELECT
  university,
  difficult_level,
  round((count(question_practice_detail.question_id))/count(distinct question_practice_detail.device_id),4)
  as avg_answer_cnt
from
  user_profile
  join question_practice_detail on user_profile.device_id=question_practice_detail.device_id
  join question_detail on question_practice_detail.question_id=question_detail.question_id
where
  user_profile.university='山东大学'
group by
  university,difficult_level